Mathematics - Circle Question with Solution | TestHub

Since QA and QB are tangents to the circle at A and $\mathrm{B},\angle OAQ=\angle OBQ=90^{\circ }$, so $\mathrm{O},\mathrm{A},\mathrm{Q},\mathrm{B}$ are concyclic on the circle with OQ as diameter: $x^2+y^2-x\alpha -y\beta =0$, with centre $\left(\frac{\alpha }{2},\frac{\beta }{2}\right)$. Let the circumcentre be $(h,k):h=\frac{\alpha }{2}$, $k=\frac{\beta }{2}\Rightarrow \alpha =2h,\beta =2k$. Also $\beta t=\alpha +2t^2$ gives $\alpha -2\beta +8=0$ for $t=2$. Substituting: $h-2k+4=0.\therefore$ locus is $x-2y+4=0$.